Calculating Impedance Using Capacitive Reactance

AI Thread Summary
To calculate the impedance of a circuit with a 29kΩ resistor and a 62mH inductor at an 80Hz frequency, the inductive reactance (XL) is determined to be 31.16Ω using the formula XL = ωL. The discussion clarifies that since this is an RL circuit, the impedance (Z) can be calculated using the formula Z = [(R^2 + XL^2)]^1/2, without needing to find capacitive reactance (XC). The relationship between voltages in the circuit is explained through Kirchhoff's voltage law, emphasizing that inductive and capacitive voltages interact vectorially. The final impedance calculation confirms that XC is not required for this specific RL circuit analysis.
PeachBanana
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Homework Statement



A 29kΩ resistor is in series with a 62mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is 80Hz .

Homework Equations



XL = ωL
Z = [(R^2 + (XL - XC)^2)] ^ 1/2


The Attempt at a Solution



R = 29,000 Ω

XL = 0.062 H (2∏ * 80 Hz)
XL = 31.16 Ω

I'm having difficulty figuring out how to find XC. XC = 1 / ωC but what is "C?"
 
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PeachBanana said:

Homework Statement



A 29kΩ resistor is in series with a 62mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is 80Hz .

Homework Equations



XL = ωL
Z = [(R^2 + (XL - XC)^2)] ^ 1/2

The Attempt at a Solution



R = 29,000 Ω

XL = 0.062 H (2∏ * 80 Hz)
XL = 31.16 Ω

I'm having difficulty figuring out how to find XC. XC = 1 / ωC but what is "C?"
Your formula for Z applies to RLC circuits. You have an RL circuit. Do you know how that formula for Z was derived? It was probably using a phasor diagram. If you understand the derivation, it's pretty easy to see what you need to do.
 
Was the formula for "Z" derived using the Pythagorean theorem? If so, I still do not see the connection to capacitance.
 
Describe how the formula was derived.
 
It looks as if it uses the following relations:

Vrms = Irms * Z
Vnaught = InaughtZ

The book also shows a phasor diagram where Vnaught is the sum vector and acts as the hypotenuse of the triangle.
 
And where does the triangle come from?
 
The triangle comes from a phasor diagram. I'm attaching an image made in paint of the one I see in the textbook. It's a tad messy, but it is readable.
 

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In your picture, I should actually point in the same direction as VR=IR.

Do you understand why the one leg is VL-VC and why you add all the voltages?
 
Are the voltages being added together to find the peak voltage source, Vnaught?
Is Vc subtracted from Vl because Vc lags the current by 90°?
 
  • #10
PeachBanana said:
Are the voltages being added together to find the peak voltage source, Vnaught?
Yes, it's Kirchoff's voltage law applied to the elements in series.

Is Vc subtracted from Vl because Vc lags the current by 90°?
And because VL leads the current by 90°. The two phasors point in opposite directions, so when you add them vector-wise, you subtract their magnitudes.

So back to your original problem, if you follow the same analysis, what do you get for the impedance?
 
  • #11
Since this is an RL circuit then

Z = [(R^2 + XL^2)]^1/2
 
  • #12
Yup, you have it.
 
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